16.5,16.6
Surface Integrals
Review :
r  u , v   x(u , v), y (u , v), z (u , v)
Parametrization of a surface:
where (u, v) lie in some region in the uv plane
surface area element: d  ru  rv
area ( S )   d =
S
 r  r
u
v
dudv
S
Ex: a sphere: r  ,     a sin( ) cos( ), a sin( ) sin( ), a cos( )  , r  r  a 2 sin( )
Surface area of the graph: z  f ( x, y ).
area(S ) 

1  f x2  f y2 dxdy
r  u, v    u, v, f (u, v) 
Example : Find the surface area of the portion of the cone
z 2  x 2  y 2 lying above the disc (x  1) 2  y 2  1.

area( S ) 
1  f x2  f y2 dxdy
z  f ( x, y )  x 2  y 2
fx 
1
2 x y
2
2
 2x 
x
x y
2
fy 
2
2
2
x
y
1 f  f  1 2

x  y 2 x2  y 2
2
x
2
y

area( S ) 

1  f x2  f y2 dxdy
area( S ) 

2dxdy  2  area( R)  2
R
y
x2  y 2

2 x2  y 2
x y
2
2

2
Surface area of a level surface : F ( x, y, z )  c
assume that we can solve for z, i.e. z  h( x, y ) with F ( x, y, h( x, y ))  c
h
differentiate implicity with respect to x : Fx +Fz   0
x
Fy
Fx
hx = 
, hy = 
Fz
Fz
Surface area element of z  h( x, y ) : d  1  hx2  hy2 dxdy
2
 Fx   Fy 
 1     
 Fz   Fz 
2
1

Fz
F
F F F =
Fz
2
x
2
y
2
z
Surface area element of F ( x, y, z )  c : d 
area( S )    d 
S

R
F
Fz
dxdy
F
dxdy
Fz
where R is the projection of S onto the xy plane.
(we need to assume S projects uniquely, i.e. no folding)
Example : Find the surface area of a spherical cap of height h
in a sphere of radius a: x 2  y 2  z 2  a 2 , a  h  z  a.
F  x2  y 2  z 2  a2
F  2 x, 2 y, 2 z
F  4 x 2  4 y 2  4 z 2  4  x 2  y 2  z 2 
 4  a  2a
2
area( S ) 

R
F
dA
Fz
F  2 a
F 2 a a


Fz
2z z
1
1
dA
 a  dA  a 
z
a2   x2  y 2 
R
R
R is the projection of the cap into the xy plane.
A disc whose boundary is a circle.
Intersection between
x 2  y 2  z 2  a 2 and z  a  h
x 2  y 2  a 2   a  h   2ah  h 2
2
Surface Area  a 
R
2
a
0
2
1
a2   x2  y 2 
2 ah  h 2

0
r
a r
2
2
dA
R: x 2  y 2  2ah  h 2
drd
u  a2  r 2
du  2rdr
 a  d    a 2  r 2 

0
0
1
2
2 ah  h 2
1
2
du  rdr
1/2
u
 du 
1/2
u
1
2
1
2
 u
 2 a   a 2   2ah  h 2   a 


 2 a   a 2  2ah  h 2  a 


 2 a    a  h   a   2 ah
Surface area of a spherical cap of height h
in a sphere of radius a is equal to 2 ah
(if h  2a, we get the whole
sphere with area 2 ah  4 a 2 )
General surface integral
  G  x, y, z  d where
S is a surface in 3-space.
S
  Gd 
S
2
2
G
1

f

f
if z  f ( x, y ) and R is the
x
y dxdy

R
projection of S onto the xy plane
  Gd =  G r  r
u
S
v
dudv where r  u, v   x(u , v), y (u , v), z (u , v)
R
describes S in parametric form.
F
 S Gd = R G Fz dxdy where F ( x, y, z )  c describes S implicitly,
and R is the projection of S onto the xy plane
Example : The surface S: y  x 2 lies in the first quadrant and is bounded by
y  0, y  4, z  0, z  3. It has mass density equal to the distance to the yz plane.
What is its total mass?
mass density   x
total mass =  xd
S
cannot project into the xy plane (why not?)
instead project into the xz plane (or yz plane): y  f ( x, z )  x 2 (or x  f ( y, z )  y )
f x  2 x, f z  0
d  1  f x2  f z2 dxdz  1  4 x 2 dxdz
total mass   x 1  4 x 2 dxdz
if 0  y  4, then 0  x  2
S
3 2
   x 1  4 x 2 dxdz
0 0
3
2
12
2 3/2 
  dz 
1 4x  



0
83
0


1
3/2
 17   1
4
Flux through a surface
Recall: Outward flux across a simple closed
curve C in the plane is
Let F  x, y, z   v be the velocity field of a fluid.
Volume of fluid flowing through
an element of surface area  per unit time
is approximated by
 compn v     v  n  d
The total volume of fluid flowing through
the surface S per unit time is called the
flux through S
flux    v  n  d
S
 F  nds
C
In a flux integral,
  F  n d
we need to choose a normal n.
A surface S is orientable if there exists a continuous unit normal vector
function n defined at each point on the surface
orientable
surface with
n and  n
upward
orientation
n
downward
orientation
n
non-orientable
surface since
n becomes  n
 1 
To find n, define the surface S by g  x, y, z   c, then n  
 g
 g 
1
1
2
2
Example : Let F   x 2 , y 2 ,z and S the paraboloid z  4  x 2  y 2 , 0  z  4.
Compute the flux of F through S, where n is the upward pointing normal.
flux    F  n  dS
g  x2  y 2  z  4
g  2 x, 2 y,1
(upward pointing normal,
positive z coordinate)
g  4 x  4 y  1
2
x3  y 3  z
flux  
1  4 x 2  4 y 2 dA
4 x2  4 y 2  1
R
2
 1 
1
n  

g

2 x, 2 y,1

2
2
4x  4 y 1
 g 
F n 
S


flux   x3  y 3   4  x 2  y 2  dA
R
2 2
x3  y 3  z
   r  cos   sin    4  r  rdrd

4x2  4 y 2  1
z  0  x2  y 2  4
 0  r  2, 0    2
3
3
3
2
0 0
2

dS  1   z x    z y  dA
2
2
   cos   sin    2r
r5
5
3
3
0
2
dS  1  4 x  4 y dA
2
2

2

r4
4

2
d
0
   cos   sin    4 d  8
3
32
5
0

 why is

2
3
 cos  d 
0
3
2
 sin
0
3

 d  0? 

  F  n d
Cancellations in Flux integrals
S
 1 
g
If S is given by g  x, y, z   0, then n  

g
Recall:
d


dxdy

gz
 | g | 
  F  n d
=

S
S
 F g   g
| g |
gz
dxdy = 
S
 F g dxdy
| gz |
If S is given by z  f ( x, y ), or f ( x, y )  z  0,
then n 
  F  n d    F   f , f
Recall: d  1  f x2  f y2 dxdy
x
S
y
 f x , f y , 1
1  f x2  f y2
, 1 dxdy
S
If S is given by r  u, v   x(u, v), y (u , v), z (u , v) , then ru  rv is normal
to the surface and hence a unit normal is n  ru  rv
Recall: d  ru  rv dudv
ru  rv
  F  n d    F   r  r  dudv
u
S
S
v